2. Infinity and Induction

2.5. Pi and e are irrational

Pi and e (Euler's number) are two numbers that occur everywhere in
mathematics. Both are irrational and in fact transcendental numbers.
In this little appendix we will prove that both numbers are irrational.
We will not show that they are also transcendental numbers. The
proofs will actually use tools from much later sections, but they
might be interesting nevertheless, and the results fit well in this
section.

First will now prove that e (Euler's number) is irrational. In
fact, it is a transcendental number, but that requires a lot of work
to prove.

Now suppose that e was rational, i.e. there are two positive
integers a and b with e = a / b.

Choose n > b. Then, using the above representation, we have:

or multiplying both sides by n! we get:

Since n > b, the left side of this equation represents an
integer, and hence n! Rn is also an integer. But
we know that

so that if n is large enough the left side is less than 1.
But then n! Rn must be a positive integer less
than 1, which is a contradiction.

Next we prove that Pi is irrational, which is a lot harder
to do. We need a preliminary result:

Lemma 2.5.1:

Define a function fn(x) = . Then this function
has the following properties:

0 < fn(x) < 1 / n! for
0 < x < 1

and
are both integers

Proof: Using the binomial theorem, we see that when the
numerator of the function is multiplied out, the lowest power
of x will be n, and the highest power
is 2n. Therefore, the function can be written as

fn(x) =

where all coefficients are integers. It is clear from this expression
that = 0 for
k < n and for k > 2n.

Also, looking at the sum more carefully, we see that

...

But this implies that
is an integer for any k.

Moreover, since

we also have

Therefore is an integer
for any k.

[ x ]

Also note that
| | <
for n large enough ( n > 2a) and any
a. Now we can prove the result of this chapter:

Proof: We will prove that
is not rational, which implies the assertion. Suppose it was rational. Then
there are two positive integers a and b with

= a / b

Define the function

Each of the factors

is an integer, by assumption on .
We also know that
and
are integers as well.
Hence, G(0) and G(1) are both integers.

Differentiating G(x) twice, we get:

The last term in this equation, the (2n+2)-th derivative, is
zero, by the properties of the function f. Adding G(x)
and G''(x) we get:

Now define another function

Then, using the above formula for
G(x) + G''(x), we have:

= =

Now we can use the second fundamental theorem of calculus to conclude:

= H(1) - H(0)
= = [ G(1) + G(0)]

Thus, the integral

is an integer. But we also know that
0 < fn(x) < 1 / n! for 0 < x < 1.
Therefore, estimating the above integral, we get

0 < <

for 0 < x < 1. Therefore, we can estimate our integral
to get

0 < <
< 1

Here we have used the fact that the last fraction approaches zero
if n is large enough. But now we have a contradiction,
because that integral was supposed to be an integer. Since there is no
positive integer less than 1, our assumption that
was rational resulted in a
contradiction. Hence, must
be irrational.

This prove is, admittedly, rather curious. Aside from the assumption
that Pi is rational (leading to the contradiction) the only
other properties of Pi that are really involved in this
proof are that

sin() = 0,
cos() = 1

and we need the defining properties of the trig. functions that

sin'(x) = cos(x), cos'(x) = - sin(x), sin(0) = 0, cos(0) = 1

That does not make the proof much clearer, but it illustrates
that some essential properties of
are indeed used along the
way, and this proof will not work for any other number.